Question

Find the value of $\csc\left[ {2{{\cot }^{ - 1}}\left( 5 \right) + {{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right]$.
A. $\dfrac{{75}}{{56}}$
B. $\dfrac{{65}}{{56}}$
C. $\dfrac{{56}}{{33}}$
D. $\dfrac{{65}}{{33}}$

Answer 1

Hint: To solve the question, we will assume that $2{\cot ^{ - 1}}\left( 5 \right) = \theta $ and ${\cos ^{ - 1}}\left( {\frac{4}{5}} \right) = \phi $. Then calculate the value of $\cos \theta ,\sin \theta ,\cos \phi $ and $\sin \phi $ from the above equations. Using these values, calculate the value of the given expression.

Formula Used:
${\cot ^{ - 1}}\left( a \right) = {\tan ^{ - 1}}\left( {\dfrac{1}{a}} \right)$
$2{\tan ^{ - 1}}\theta = {\tan ^{ - 1}}\left( {\dfrac{{2\theta }}{{1 - {\theta ^2}}}} \right)$
${\sec ^2}\theta = 1 + {\tan ^2}\theta $
$\sec \theta = \dfrac{1}{{\cos \theta }}$
${\sin ^2}\theta = 1 - {\cos ^2}\theta $
$\csc\alpha = \dfrac{1}{{\sin \alpha }}$
$\sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta $

Complete step by step solution:
Given trigonometric expression is
$\csc\left[ {2{{\cot }^{ - 1}}\left( 5 \right) + {{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right]$
Converting ${\cot ^{ - 1}}$ into ${\tan ^{ - 1}}$ by using the formula ${\cot ^{ - 1}}\left( a \right) = {\tan ^{ - 1}}\left( {\dfrac{1}{a}} \right)$.
$ = \csc\left[ {2{{\tan }^{ - 1}}\left( {\dfrac{1}{5}} \right) + {{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right]$
Now applying $2{\tan ^{ - 1}}\theta = {\tan ^{ - 1}}\left( {\dfrac{{2\theta }}{{1 - {\theta ^2}}}} \right)$
$ = \csc\left[ {{{\tan }^{ - 1}}\left( {\dfrac{{2 \cdot \dfrac{1}{5}}}{{1 - {{\left( {\dfrac{1}{5}} \right)}^2}}}} \right) + {{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right]$
$ = \csc\left[ {{{\tan }^{ - 1}}\left( {\dfrac{{\dfrac{2}{5}}}{{1 - \dfrac{1}{{25}}}}} \right) + {{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right]$
$ = \csc\left[ {{{\tan }^{ - 1}}\left( {\dfrac{{10}}{{24}}} \right) + {{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right]$
$ = \csc\left[ {{{\tan }^{ - 1}}\left( {\dfrac{5}{{12}}} \right) + {{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right]$
Now let ${\tan ^{ - 1}}\left( {\dfrac{5}{{12}}} \right) = \theta $ ……(1) and ${\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right) = \phi $ …….(2)
Now we will calculate $\sin \theta ,\cos \theta $ from ${\tan ^{ - 1}}\left( {\dfrac{5}{{12}}} \right) = \theta $
${\tan ^{ - 1}}\left( {\dfrac{5}{{12}}} \right) = \theta $
$ \Rightarrow \left( {\dfrac{5}{{12}}} \right) = \tan \theta $
Apply formula if $\tan \theta = \dfrac{a}{b}$, then $\sin \theta = \dfrac{a}{{\sqrt {{a^2} + {b^2}} }}$
$\sin \theta = \dfrac{5}{{\sqrt {{5^2} + {{12}^2}} }}$
$ \Rightarrow \sin \theta = \dfrac{5}{{\sqrt {25 + 144} }}$
$ \Rightarrow \sin \theta = \dfrac{5}{{13}}$
Apply formula if $\tan \theta = \dfrac{a}{b}$, then $\cos \theta = \dfrac{b}{{\sqrt {{a^2} + {b^2}} }}$
$\cos \theta = \dfrac{{12}}{{\sqrt {{5^2} + {{12}^2}} }}$
$ \Rightarrow \cos \theta = \dfrac{{12}}{{\sqrt {25 + 144} }}$
$ \Rightarrow \cos \theta = \dfrac{{12}}{{13}}$
Now we will calculate $\sin \phi ,\cos \phi $ from ${\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right) = \phi $
${\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right) = \phi $
$ \Rightarrow \left( {\dfrac{4}{5}} \right) = \cos \phi $
Putting $\cos \phi = \left( {\dfrac{4}{5}} \right)$ in ${\sin ^2}\phi = 1 - {\cos ^2}\phi $
${\sin ^2}\phi = 1 - {\left( {\dfrac{4}{5}} \right)^2}$
$ \Rightarrow {\sin ^2}\phi = 1 - \dfrac{{16}}{{25}}$
$ \Rightarrow {\sin ^2}\phi = \dfrac{9}{{25}}$
$ \Rightarrow \sin \phi = \dfrac{3}{5}$
Now we will put ${\tan ^{ - 1}}\left( {\dfrac{5}{{12}}} \right) = \theta $ ……(1) and ${\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right) = \phi $ in $\csc\left[ {{{\tan }^{ - 1}}\left( {\dfrac{5}{{12}}} \right) + {{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right]$
$\csc\left[ {{{\tan }^{ - 1}}\left( {\dfrac{5}{{12}}} \right) + {{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right]$
$ = \csc\left[ {\theta + \phi } \right]$
Convert $\csc$ into $\sin $
$ = \dfrac{1}{{\sin \left[ {\theta + \phi } \right]}}$
Now applying $\sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta $
$ = \dfrac{1}{{\sin \theta \cos \phi + \cos \phi \sin \theta }}$
Now putting the values $\sin \theta ,\cos \theta ,\sin \phi ,$ and $\cos \phi $
$ = \dfrac{1}{{\sin \theta \cos \phi + \cos \theta \sin \phi }}$
$ = \dfrac{1}{{\dfrac{5}{{13}} \cdot \dfrac{4}{5} + \dfrac{{12}}{{13}} \cdot \dfrac{3}{5}}}$
$ = \dfrac{1}{{\dfrac{{20}}{{65}} + \dfrac{{36}}{{65}}}}$
$ = \dfrac{1}{{\dfrac{{20 + 36}}{{65}}}}$
$ = \dfrac{{65}}{{56}}$

Option ‘B’ is correct

Note: To solve this type of problem a student must have deep knowledge about the inverse function and trigonometry ratios. In the given question we need to assume the value of ${\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right)$ and $2{\cot ^{ - 1}}\left( 5 \right)$ to get the value of $\cos ec\left[ {2{{\cot }^{ - 1}}\left( 5 \right) + {{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right]$.